For the reaction :
`4Al+ 3O_(2) + 6H_(2)O+ 4OH^(-) rightarrow4A1(OH)_(4)^(-) E^(@)_(cell) = 2.73V`.
Standard Gibb's free energy change for the reaction is :

Options

A. `-3.16xx10^(3) kJ`
B. `-0.7 6xx10^(3) kJ`
C. `-0.263xx10^(3) kJ`
D. `+0.263xx10^(3) kJ`

Explanation

To find the standard Gibbs free energy change (ΔG°) for the given reaction, we can use the formula:

\[ \Delta G° = -nFE°_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E°_{cell} \) = standard cell potential (given as \( 2.73 \, V \))

### Step 1: Determine the number of electrons transferred (n)
In the reaction:

\[ 4Al + 3O_2 + 6H_2O + 4OH^- \rightarrow 4Al(OH)_4^- \]

We need to find out how many electrons are involved in the oxidation of aluminum. The oxidation state of aluminum (Al) changes from 0 (in elemental form) to +3 in \( Al(OH)_4^- \).

For each aluminum atom, the change in oxidation state is:
\[ 0 \rightarrow +3 \quad \text{(which means 3 electrons are lost)} \]

Since there are 4 moles of aluminum reacting, the total number of electrons transferred is:
\[ n = 4 \times 3 = 12 \, \text{electrons} \]

### Step 2: Substitute values into the Gibbs free energy equation
Now we can substitute \( n \), \( F \), and \( E°_{cell} \) into the Gibbs free energy equation:

\[ \Delta G° = -nFE°_{cell} \]
\[ \Delta G° = -12 \times 96500 \, C/mol \times 2.73 \, V \]

### Step 3: Calculate ΔG°
Calculating the above expression:

\[ \Delta G° = -12 \times 96500 \times 2.73 \]
\[ \Delta G° = -12 \times 263295 \]
\[ \Delta G° = -3159540 \, J/mol \]

To convert joules to kilojoules, we divide by 1000:

\[ \Delta G° = -3159.54 \, kJ/mol \approx -3.16 \times 10^3 \, kJ/mol \]

### Final Answer
Thus, the standard Gibbs free energy change for the reaction is:

\[ \Delta G° \approx -3.16 \times 10^3 \, kJ/mol \]

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